解:令u=2x+3y,故du/dx=2+3dy/dx=2+3(u+4)/(2u+5) 即d/dx=(7u+22)/(2a+5),7u+22=0时,得特解14x+21y+22=0, 7+22≠0时,分离变量得②2-9/(7+2)]d=7dx,积分得 2u-9/7ln(7u+22)/d]=7x,代回原变量整理得通解 7(2y-x)-3ln(14x+21y+2)/d=0.另一形式为 14x+21y+22=cexp(7(2y-x)/3),特解14x+21y+22=0包含在通 解的后一形式中 8)dy/dx=(x+1)2+(4y+1)2+8xy+1 解:可见dy/dx=(x+4y+1)2+2,故令t=x+4y+1,从而 du/dx=1+4dy/dx=1+4(2+2)=42+9,即du/d=4n2+9,分离 变量得34(2u)/(42+9)=6dx,积分得: arctan(2u/3)=6x+c代回原 变量整理得通解 arctan(2x+8y+2)/3)=6x+c. dx=(y6-2x2)/(2xy5+x2y2) 解:将原方程化为d(y3)/dx=3(y3)2-2x21/(2xy3+x2),令v=y3 得齐次方程d/dx=3(2-2x2)/(x+x2),故令=xu,对x求 导得d/dx=u+rdu/dx=3(2-2)/(2u+1).即得变量分离方程 rda/dx=(u-3)(u+2)/(2u+1,分离变量得 (x-3)+3/(+2)du=5dx/x.积分得7ln|u-3l+3lnl+2|=5ln(cr) 代回原变量得通解(y3-3x)7(y3+2m)3=c15.(注:对应于u=3及 u=-2的特解包含在通解中) 10)dy/dx=(2x3+3xy2+x)/(3x2y+2y3-y) 解:将原方程化为 d(y2)/d(x2)=[2(x2-1)+3(y2+1)/3(x2-1)+2(y2+1)从而可令 u=x2-1,v=y2+1,原方程化为齐次方程dv/du=(2u+30)(3u+2a), 得变量分离方程w/u=2(1-2)(2m+3).分离变量得3 令U=,对u求导得,dv/da=+udu/da=(3u+2)/(2u+3).即 1/(+1)-5/(-1)da=4du/u,积分得 lnlu+1-5lmlu-l}=4lnla+c,代回原变量整理得通解 (y2-x2+2)5=c(x2+y2).(注:对应于v=1的特解包含在通解中) 用常数变易公式求解下列(可化为)线性方程或 Bernoulli方 程的通解或初值问题 1)dy/dr=y+sin: c 解:取线性齐次方程dy/dx-y=0的一个特解hx)=exp(x),应 用常数变易公式得 y=exp(a)c+sin r exp(a)d r= cexp(a)-(sin c +cos )/2 2)d r/dt exp(2t)-3c 解:取对应的齐次方程的一个特解为h(t)=exp(-3t),应用常数 变易公式得:x=exp(-3)c+∫ep(5t)d]=cexp(-3t)+exp(2t)/5 3 dy/dc-ny/a=rnexp(a) 答:y=x"(c+exp(x) 4)dy/dx+(1-2r)y/x2-1=0 解:取对应的齐次方程的一个特解h(x)=x2exp(1/x),应用常数 变易公式得y=x2exp(1/a)+∫exp(-1/x)d(-1/x)]=x2 [cexp(1/x)+1] 5 dy/dr=ytan r +cos c: U u = 2x + 3y, V du/dx = 2 + 3dy/dx = 2 + 3(u + 4)/(2u + 5), du/dx = (7u + 22)/(2u + 5), 7u + 22 = 0 , v? 14x + 21y + 22 = 0, 7u + 22 6= 0 ,
:*v [2 − 9/(7u + 22)] du = 7 dx,, )
v, 2u − 9/7 ln[(7u + 22)/c] = 7x, 7R;*|v= 7(2y − x) − 3 ln[(14x + 21y + 22)/c] = 0. WO/1 14x + 21y + 22 = c exp(7(2y − x)/3), ? 14x + 21y + 22 = 0 cTA= O/X. 8) dy/dx = (x + 1)2 + (4y + 1)2 + 8xy + 1 : RF dy/dx = (x + 4y + 1)2 + 2, VU u = x + 4y + 1, uW du/dx = 1 + 4dy/dx = 1 + 4(u 2 + 2) = 4u 2 + 9,  du/dx = 4u 2 + 9,
: *v3d(2u)/(4u 2 + 9) = 6 dx, )
v: arctan(2u/3) = 6x + c, 7R; *|v= arctan((2x + 8y + 2)/3) = 6x + c. 9) dy/dx = (y 6 − 2x 2 )/(2xy5 + x 2y 2 ) : ~;+1 d(y 3 )/dx = 3[(y 3 ) 2 − 2x 2 ]/(2xy3 + x 2 ), U v = y 3 , v01 dv/dx = 3(v 2 − 2x 2 )/(2xv + x 2 ), VU v = xu, Z x ] ^v dv/dx = u + xdu/dx = 3(u 2 − 2)/(2u + 1). v*
: xdu/dx = (u − 3)(u + 2)/(2u + 1),
:*v [7/(u−3)+3/(u+2)] du = 5dx/x. )
v 7 ln |u−3|+3 ln |u+2| = 5 ln(cx). 7R;*v= (y 3 − 3x) 7 (y 3 + 2x) 3 = cx15. (}: Z5z u = 3 g u = −2 ? cTA= X). 10) dy/dx = (2x 3 + 3xy2 + x)/(3x 2y + 2y 3 − y). : ~;+1 d(y 2 )/d(x 2 ) = [2(x 2 − 1) + 3(y 2 + 1)]/[3(x 2 − 1) + 2(y 2 + 1)]. uWRU u = x 2 − 1, v = y 2 + 1, ;+101dv/du = (2u + 3v)(3u + 2v), U v = uw, Z u ]^v, dv/du = w + udw/du = (3w + 2)/(2w + 3).  v*
: udw/du = 2(1 − w 2 )(2w + 3).
:*v [1/(w + 1) − 5/(w − 1)] dw = 4du/u, )
v: ln |w + 1| − 5 ln |w − 1| = 4 ln |u| + c, 7R;*|v= (y 2 − x 2 + 2)5 = c(x 2 + y 2 ). (}: Z5z w = 1 ? cTA= X) 3. 9*-./] *+ (R+1 O Bernoulli = O@: 1) dy/dx = y + sin x : k01 dy/dx − y = 0 N? h(x) = exp(x), 5 9*-./v: y = exp(x)[c + R sin x exp(−x) dx] = c exp(x) − (sin x + cos x)/2. 2) dx/dt = exp(2t) − 3x : kZ5 01 N? 1 h(t) = exp(−3t), 59 *-./v: x = exp(−3t)[c + R exp(5t) dt] = c exp(−3t) + exp(2t)/5. 3) dy/dx − ny/x = x n exp(x) 2: y = x n (c + exp(x)). 4) dy/dx + (1 − 2x)y/x2 − 1 = 0 : kZ5 01 N? h(x) = x 2 exp(1/x), 59 *-./v y = x 2 exp(1/x)[c+ R exp(−1/x) d(−1/x)] = x 2 [c exp(1/x)+1]. 5) dy/dx = y tan x + cos x 5